OK, that helps. Do you understand what that means? In other words do you know the meaning of gain bandwidth product? We need to have a baseline of whether your trouble is that you don't understand the definition, or if you just don't know how to apply the definition to that particular problem.

Your above information is still a little vague though. So is the 100 Hz frequency referenced to a particular inverting opamp, or to the one in your circuit? Or, is it referenced to the OPAMP itself, without any components (open loop)?

The input resistors attenuate the input 101 times then the feedback resistors cause the opamp to amplify 1001 times.
The result is an amplification of only about 9.9 times. The output will probably be very noisy (hiss).

An audio opamp (something better and newer than a lousy old 741 opamp) that is open loop has a gain of 33,000 to about one million at 100Hz.

The input resistors attenuate 101 times and the feedback causes 1001 times gain.
So the result is a gain of about only -9.9 times.

Click to expand...

How are you defining your feedback portion? If it's the two resistors 1M and 1K, then this is an inverting amplifier configuration and the gain of that portion is -R2/R1=-1000.

Looking at the whole circuit, you can make a Thevenin equivalent source of Vi*10/1010 and a source resistance of 10 in parallel with 1000, which is about 9.9 Ohms. Or Vi/101 and 9.9 ohm source resistance. Now the effective input resistance on the inverting amplifier (looking in from the ideal Thevenin voltage of Vi/101) is 1009.9 ohms due to the source resistance, and the net gain of the inverting Opamp is -1000000/1009.9=-990. Now combine this with the attenuated voltage and the net gain is -990/101 which is about equal to 9.8. Isn't it?

Alternatively, you can analyze the full circuit with equations and you come up with Av=-R3*R2/(R4*R1+R3*(R4+R1)) where R2 and R1 are the OPAMP feedback resistors 1M and 1K, respectively; and R4 and R3 are the voltage divider resistors 1000 and 10, respectively. Again, the gain is about -9.8.

Another way of calculating the signal gain is to note that Ra, Rb and R1 form a T network. Use the delta-wye transformation to convert it into a pi network. Then the shunt resistors of the pi network can be ignored because the one across the input voltage has no effect if the input voltage source has zero output impedance, and the shunt across the - input of the opamp has no effect because that terminal is a virtual ground.

We are left with the series element of the pi network, which is 102000 ohms. Then the signal gain is 1000000/102000 = 500/51 = 9.80392

Another way of calculating the signal gain is to note that Ra, Rb and R1 form a T network. Use the delta-wye transformation to convert it into a pi network. Then the shunt resistors of the pi network can be ignored because the one across the input voltage has no effect if the input voltage source has zero output impedance, and the shunt across the - input of the opamp has no effect because that terminal is a virtual ground.

We are left with the series element of the pi network, which is 102000 ohms. Then the signal gain is 1000000/102000 = 500/51 = 9.80392

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Very elegant method !

I'm still a little confused on the OPs question. I have the gist of what he's asking, but the precise question and the precise information he was given is not fully clear to me.

For example, "What are the two estimates one can make?". I'm not sure what this means. It seems we need to have the OPAMPs gain-bandwidth product to make any estimates, and once we have that, shouldn't we just have one estimate for the entire circuit GB product?

I don't know what is wanted either. The only thing I can think of is related to the strange presence of the T network at the input. Why would a person have that T network instead of a single 102000 ohm resistor?

I think the noise gain of the circuit is different with the T network rather than with a single resistor.

I'm still a little confused on the OPs question. I have the gist of what he's asking, but the precise question and the precise information he was given is not fully clear to me.

For example, "What are the two estimates one can make?". I'm not sure what this means. It seems we need to have the OPAMPs gain-bandwidth product to make any estimates, and once we have that, shouldn't we just have one estimate for the entire circuit GB product?

Maybe I'm missing something?

Click to expand...

That is all that was asked. I left off no other information. Not an important question now.

sometimes I think that professors need to actually look at a databook once in a while so that their questions are not quite so open to variable interpretation (or generate more questions). Maybe that was the intent?